Question: $ \int_{-2}^2 \int_{1 - \sqrt{4 - x^2}}^{1 + \sqrt{4 - x^2}} dy \, dx$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{-1}^3 \int_{-\sqrt{4 - y^2}}^{\sqrt{4 - y^2}} dx \, dy$ (Choice B) B $ \int_0^2 \int_{-\sqrt{1 - (y - 1)^2}}^{\sqrt{1 - (y - 1)^2}} dx \, dy$ (Choice C) C $ \int_{-1}^3 \int_{-\sqrt{4 - (y - 1)^2}}^{\sqrt{4 - (y - 1)^2}} dx \, dy$ (Choice D) D $ \int_0^2 \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} dx \, dy$
Explanation: The first step whenever we want to switch bounds is to sketch the region of integration that we're given. Here, we see $-2 < x < 2$ and $1 - \sqrt{4 - x^2} < y < 1 + \sqrt{4 - x^2}$. Therefore: ${1}$ ${2}$ ${3}$ ${\llap{-}1}$ ${\llap{-}2}$ ${\llap{-}3}$ ${1}$ ${2}$ ${3}$ ${4}$ ${\llap{-}1}$ ${\llap{-}2}$ $y$ $x$ Because we're switching bounds to $dx \, dy$, we need to start with numeric bounds for $y$. We see that $-1 < y < 3$. Now we can define $x$ in terms of $y$. $-\sqrt{4 - (y - 1)^2} < y < \sqrt{4 - (y - 1)^2}$ We want to pay attention especially to how this $x$ bound works at the edge of the $y$ bound. For example, at $y = 3$, the $x$ bound makes $x = 0$ as expected. In conclusion, the double integral after switching bounds is: $ \int_{-1}^3 \int_{-\sqrt{4 - (y - 1)^2}}^{\sqrt{4 - (y - 1)^2}} dx \, dy$